Explanation Theorem used: Divergence theorem, ∬ S F . d S = ∭ R d i v ( F ) d V Calculation: Now calculating ∬ S F . d S , for this we use parametrization of cylinder surface Surface S : ϕ ( θ , t ) = ( t cos θ , t sin θ , t 2 ) , 0 ≤ t ≤ 2 , 0 ≤ θ ≤ 2 π Outward normal is given by, n = | i j k − t sin θ t cos θ 0 cos θ sin θ 2 t | = 〈 2 t 2 cos θ , 2 t 2 sin θ , − t 〉 F ( ϕ ( θ , t ) ) . n = 〈 t cos θ , 0 , 0 〉 . 〈 2 t 2 cos θ , 2 t 2 sin θ , − t 〉 = 2 t 3 cos 2 θ ∬ S F . d S = ∫ 0 2 π ∫ 0 2 F ( ϕ ( θ , t ) ) . n d t d θ = ∫ 0 2 π ∫ 0 2 2 t 3 cos 2 θ d t d θ = ( ∫ 0 2 π cos 2 θ d θ ) ( ∫ 0 2 2 t 3 d t ) = 1 2

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Chapter 18.3, Problem 4E

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The verification of divergence theorem for the vector field and $F (x, y, z) = (x, 0, 0)$ the region of $x^{2} + y^{2} \leq z \leq 4$ .

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Chapter 18.3, Problem 3E

Chapter 18.3, Problem 5E

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Calculate the divergence of the vector field:

V=(y,2y,0) and w=(x-y,x+y,0)

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The verification of divergence theorem for the vector field and F ( x , y , z ) = ( x , 0 , 0 ) the region of x 2 + y 2 ≤ z ≤ 4 .

Solution Summary: The author explains the verification of the divergence theorem for the vector field and the region of x 2 + y 2.

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The verification of divergence theorem for the vector field and $F (x, y, z) = (x, 0, 0)$ the region of $x^{2} + y^{2} \leq z \leq 4$ .

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Chapter 18 Solutions

The verification of divergence theorem for the vector field and F ( x , y , z ) = ( x , 0 , 0 ) the region of x 2 + y 2 ≤ z ≤ 4 .

Solution Summary: The author explains the verification of the divergence theorem for the vector field and the region of x 2 + y 2.

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The verification of divergence theorem for the vector field and F(x,y,z)=(x,0,0) the region of x2+y2≤z≤4.

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Chapter 18 Solutions

The verification of divergence theorem for the vector field and $F (x, y, z) = (x, 0, 0)$ the region of $x^{2} + y^{2} \leq z \leq 4$ .